Tangents to Implicit Curves, Horizontal and Vertical
- Find a tangent line to a curve defined implicitly
- Locate points where an implicit curve has a horizontal or vertical tangent
- Recognize why an implicitly defined relation can have two tangent slopes at one x
Why implicit differentiation is not a separate technique
Implicit differentiation is the chain rule applied to y, and framing it that way removes most of the confusion. Because y is a function of x, differentiating any expression in y with respect to x picks up a factor of dy/dx: d/dx[y³] = 3y²·(dy/dx), exactly as d/dx[u³] = 3u²·(du/dx). Differentiate both sides of the equation term by term, then collect every term containing dy/dx on one side and everything else on the other, and factor. The result is generally a formula for dy/dx in terms of both x and y, which is not a defect — it is what an implicit relation gives you, and it is why you need a point rather than just an x value to get a number.
Horizontal and vertical tangents from the fraction
Once dy/dx is written as a single fraction, the two tangent questions read directly off it. A horizontal tangent needs dy/dx = 0, which happens where the numerator is zero and the denominator is not. A vertical tangent needs dy/dx to be undefined, which happens where the denominator is zero and the numerator is not. Two cautions the exam exploits. First, any candidate point must actually lie on the curve — solve the numerator or denominator condition together with the original equation, not separately. Second, if numerator and denominator vanish at the same point, neither conclusion follows and the point requires separate analysis; on a circle-like curve this typically signals a singular point.
One x, two slopes — and why that is fine
An implicit relation need not be a function. The circle x² + y² = 25 has two points at x = 3, namely (3, 4) and (3, −4), and implicit differentiation gives dy/dx = −x/y, which evaluates to −3/4 at the first and +3/4 at the second. Students sometimes treat this as an error; it is the whole point of the method. The formula for dy/dx depends on y precisely so that it can distinguish the branches. This is also the practical reason a question about an implicit curve always supplies a point rather than an x-coordinate: without y, the slope is genuinely ambiguous.
For the curve x² + xy + y² = 7, find dy/dx, then find the equation of the tangent line at the point (1, 2).
- 1.Differentiate both sides with respect to x. The x² term gives 2x. The xy term needs the product rule: 1·y + x·(dy/dx). The y² term gives 2y·(dy/dx). The constant gives 0.
- 2.So 2x + y + x(dy/dx) + 2y(dy/dx) = 0.
- 3.Collect the dy/dx terms: (x + 2y)(dy/dx) = −2x − y, so dy/dx = −(2x + y)/(x + 2y).
- 4.Verify (1, 2) is on the curve: 1² + (1)(2) + 2² = 1 + 2 + 4 = 7. It is.
- 5.Substitute: dy/dx = −(2(1) + 2)/((1) + 2(2)) = −4/5.
- 6.Point-slope form: y − 2 = (−4/5)(x − 1).
The xy term requires the product rule, and forgetting it is the single most common implicit-differentiation error. d/dx[xy] = y + x(dy/dx) — two terms, not one. Any product of x with y, or of y with a function of x, behaves the same way.
For a curve with dy/dx = (2x − y)/(x + 3y), a vertical tangent could occur where
On the circle x² + y² = 25, dy/dx = −x/y. At the two points where x = 3, the slopes are
Answer the 2 checkpoints as you read.
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